《01字典树》

只有01的字典树。

大概就是(偷图.jpg)

洛谷：P4551 最长异或路径

维护每个点到根的异或和。

然后我们对于每个点的异或和去找最优的字典树走法。

从最高位开始往另一位走。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tu;
const int N = 1e5 + 5;
const int M = 1e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int n;
struct Node{int to,dis;};
vector<Node> G[N];
int tree[N * 31][2],tot = 0,a[N];
void insert(int val) {
    int x = 0;
    for(int i = 30;i >= 0;--i) {
        int g = ((val >> i) & 1);
        if(tree[x][g] == 0) tree[x][g] = ++tot;
        x = tree[x][g];
    }
}
int query(int val) {
    int mx = 0,x = 0;
    for(int i = 30;i >= 0;--i) {
        int g = (val >> i) & 1;
        if(tree[x][g ^ 1] != 0) {
            mx += (1 << i);
            x = tree[x][g ^ 1];
        }
        else x = tree[x][g];
    }
    return mx;
}
int ans = 0;
void dfs(int u,int fa,int val) {
    if(u != 1) {
        insert(val);
        ans = max(ans,val);
        a[u] = val;
    }
    for(auto v : G[u]) {
        if(v.to == fa) continue;
        dfs(v.to,u,val ^ v.dis);
    }
}
void solve() {
    n = read();
    for(int i = 1;i < n;++i) {
        int u,v,w;u = read(),v = read(),w = read();
        G[u].push_back(Node{v,w});
        G[v].push_back(Node{u,w});
    }
    dfs(1,0,0);
    for(int i = 2;i <= n;++i) ans = max(ans,query(a[i]));
    printf("%d
",ans);
} 
int main() {    
    solve();
    system("pause");
    return 0;
}